continuity for TVS I have some problem as following：
Let $X$ be the space of all sequences of scalars. 
Define addition and scalar (= Real numbers) multiplication in the usual way. If $x_n$, $y_n\in X$, define 
$\displaystyle d(x, y) = \sum_{n=1}^{\infty} 2^{-n}\frac{|x_n - y_n|}{1+|x_n - y_n|}$.
It's already known $d$ is a metric, but I have some trobule to show the continuity of addition and continuity of scalar multiplication (i.e. to show that $X$ is topological vector space (TVS)). 
I'm trying to estimate by inequality (e.g. Triangle inequality), but there is no way. How can I do this gap？ Thanks.
 A: Personally, I find that the metric obscures whats going on and that this is easier to see once we can get rid of the metric altogether and appeal to a more general topological result. Let me show how you can do this. 
First notice that the space of real sequences with the metric $d$ is nothing other than the product space $\mathbb{R}^\mathbb{N}$ equipped with the product topology (where $\mathbb{R}$ is viewed as an $\mathbb{R}$-topological vector space for its usual metric).
So now it suffices to know that a product of topological vector spaces is again a topological vector space (for the obvious addition and scalar multiplication). For this suppose $X_i$ ($i \in \Lambda$) are topological vector spaces and $X = \prod_{i \in \Lambda} X_i$. Then we can identify (via the obvious homeomorphism) $X \times X \cong \prod_{i \in \Lambda} X_i \times X_i$.
If $\oplus:X \times X \to X$ is addition in $X$ and $\oplus_i: X_i \times X_i \to X_i$ is addition in $X_i$ then (under the above identification) we have $\oplus = \prod_{i \in \Lambda} \oplus_i$. Then, since $\oplus_i$ is continuous for each $i$, it suffices to show that a product of continuous maps is continuous. This is a standard topological result. Indeed, if $f_i: E_i \to F_i$ are continuous then for any basic open set $\prod U_i \subseteq F = \prod F_i$ (where $U_i = E_i$ for all but finitely many $i$), we have 
$$\bigg(\prod f_i\bigg)^{-1}\bigg(\prod U_i\bigg) = \prod f_i^{-1}(U_i)$$
which is a basic open set in $\prod E_i$. 
The same idea will also give continuity of scalar multiplication with a tiny bit more work. We can identify$\prod (\mathbb{R} \times X_i) \cong \bigg(\prod \mathbb{R} \bigg) \times X$ in the obvious way and then note that $\otimes: \mathbb{R} \times X \to X$ is then a restriction of the product map $\prod  \otimes_i$ (up to a composition with the obvious embedding $\mathbb{R} \hookrightarrow \prod \mathbb{R}$ given by $\lambda \mapsto (\lambda, \lambda, \lambda, \dots)$).
A: Showing that $d$ is a metric is not entirely trivial. But if, as you say, you already know that then proving addition is continuous is simple, from the triangle  inequality. Note first that it's clear that $d$ is translation invariant, meaning  that $d(x+z,y+z)=d(x,y)$. So $$\begin{align}
d(x+y,a+b)&=d((x-a)+(y-b),0)
\\&\le d((x-a)+(y-b),(y-b))+d(y-b,0)
\\&=d(x-a,0)+d(y-b,0)
\\&=d(x,a)+d(y,b).\end{align}$$So if $d(x-a)<\epsilon/2$ and $d(y-b)<\epsilon/2$ then $d(x+y,a+b)<\epsilon$.
